From the code presented for Tower of Hanoi problem: # We have 3 towers A, B and C or 3 Stacks A = [14,13,12,11,10,9,8,7,6,5,4,3,2,1] B = [] C = [] # According to our rules, if we want to move all disks 3,2 and 1 from A to C, we first move two disks from A to B. # We then move the base disk from A to C and then the stack of two from B to C count = 0 def towers_of_hanoi(A,B,C,n): global count if n == 1: disk = A.pop() C.append(disk) count +=1 else: towers_of_hanoi(A,C,B,n-1) towers_of_hanoi(A,B,C,1) towers_of_hanoi(B,A,C,n-1) return count Question 1: If you set n=3 while keeping 14 disks in peg A this routine works and produces following results: towers_of_hanoi(A,B,C,3) = 7 A = [14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4] B = [] C = [3, 2, 1] There is no constraint on 'n' in the code. How does Python know to stop after moving 3 disks from A to C? Question 2: Again, since there is no constraints on 'n' why doesn't it work for n<0? If you set n<0 Python throws an error.

Hey 이삭, Thank you for your question! The constraint on n is given by the base case, namely n == 1. Therefore, it knows to not go below n = 1. Hope this helps! Kind regards, 365 Hristina

Hi Hristina, thanks for responding to my question. I'm confused by the presence of the 'else' statement. It appears any value for n other than n==1, including n<0 would fall under the 'else' statement in which towers_of_hanoi(A,B,C,1) would always be executed. In fact, this would be an infinite loop until the stack A gets exhausted and becomes empty. What am I missing? - I would appreciate if you could walk me through step by step understanding this code. Thanks once again for your help. Best Isaac Kim